Home » Mean Value Theorem

Prove the integral from 0 to x of sin t /(t2 + 1) is positive for positive x

Prove that for all we have

Before proceeding with the proof, we recall the second mean-value theorem for integrals (Theorem 5.5 on p. 219 of Apostol). For a continuous function on the interval if has a continuous derivative which never changes sign on the interval then there exists a such that

Proof. Now, we want to apply the mean-value theorem above with and . Since is continuous everywhere , it is continuous on any interval . Then,

is continuous for all (so, in particular, for all ). Furthermore, since for all we have that for all . Thus, is continuous and never changes sign in any interval . Therefore, we can apply the mean-value theorem to conclude there exists a for any such that

But since for all we know for any and . Hence,

Application of the mean-value theorem for integrals

1. Let be a function with second derivative continuous and nonzero on an interval . Furthermore, let be a constant such that

Use the second mean-value theorem for integrals (Theorem 5.5 in Apostol) to prove the inequality

2. If prove that

1. Proof. Since we have the assumption that for all we may divide by , to obtain

Then, to apply the second mean-value theorem for integrals (Theorem 5.5 of Apostol) we define functions

The function is continuous since is a composition of continuous functions (we know is continuous since it is differentiable) and is continuous (again, it is differentiable since exists and is continuous by assumption). Then the product of continuous function is also continuous, which establishes that is continuous. We also know that meets the conditions of the theorem since it has derivative given by

This derivative is continuous since and are continuous and is nonzero. Furthermore, this derivative does not change since on since is nonzero on (and by Bolzano’s theorem we know that a continuous function that changes sign must have a zero). Therefore, we can apply the second mean-value theorem:

2. Proof. Using part (a), we take , giving us

Thus, is continuous and never changes sign. Furthermore, (where is a given constant) and since . Thus,

Find a function with continuous second derivative satisfying given conditions

In each of the following cases find a function with continuous second derivative satisfying the given conditions.

1. for all , , and .
2. for all , , and .
3. for all , , and for all .
4. for all , , and for all .

1. There can be no function meeting all of these conditions since implies is an increasing function (since its derivative, , is positive). But then contradicts that is increasing.
2. Let . Then

Furthermore, for all .

3. There can be no function meeting all of these conditions. Again, for all implies that is increasing for all . Therefore, implies for all . Then, by the mean-value theorem, we know that for any there exists some such that

Now, choose . Then, , contradicting that for all .

4. We’ll define piecewise as follows

Then, we can take the derivative of each piece (and see that they are equal, so the derivative is defined at )

Taking the derivative again we find

Thus, for all and . Furthermore, for we have

Prove some inequalities using the mean value theorem

Prove the following inequalities using the mean value theorem:

1. .
2. , for .

1. Proof. Define and . Then and are continuous and differentiable everywhere so we may apply the mean value theorem. We obtain

The final step follows since for all

2. Proof. Let , , then and . So, by the mean-value theorem we have there exists a such that,

But, since is an increasing function on the positive real axis, and we have we know

Further, since and is positive we can multiply all of the terms in the equality by without reversing inequalities to obtain,

Therefore, substituting from above we obtain the requested inequality:

Prove properties about the zeros of a polynomial and its derivatives

Consider a polynomial . We say a number is a zero of multiplicity if

where .

1. Prove that if the polynomial has zeros in , then its derivative has at least zeros in . More generally, prove that the th derivative, has at least zeros in the interval.
2. Assume the th derivative has exactly zeros in the interval . What can we say about the number of zeros of in the interval?

1. Proof. Let denote the distinct zeros of in and their multiplicities, respectively. Thus, the total number of zeros is given by,

By the definition given in the problem, if is a zero of of multiplicity then

Taking the derivative (using the product rule), we have

Thus, again using the definition given in the problem, is a zero of of multiplicity .
Next, we know from the mean-value theorem for derivatives, that for distinct zeros and of there exists a number (assuming, without loss of generality, that ) such that . Hence, if has distinct zeros, then the mean value theorem guarantees numbers such that . Thus, has at least:

By induction then, the th derivative has at least zeros.

2. If the th derivative has exactly zeros in , then we can conclude that has at most zeros in .

Prove an alternate expression for the mean-value formula

Prove that the expression

is an equivalent form of the mean-value theorem.
Find the value of in terms of and when:

1. ;
2. .

For parts (a) and (b) keep fixed with and find the limit of as tends to 0.

Proof. If is continuous on and differentiable on , then by the mean-value theorem we have

Letting and for some (since ), we have

(This follows since from our definitions, is the distance from . Then, since is somewhere in the interval its value must be plus some portion of the distance to . This portion is then , which is how we know .) Substituting and and ,

where and

Now for parts (a) and (b).

1. If , we have , so,

2. If , we have . So,

Show that x^2 = x*sin x + cos x for exactly two real numbers x

Consider the equation

Show that there are two values of such that the equation is satisfied.

Proof. Let . (We want then to find the zeros of this function since these will be the points that .) Then,

Since for any (since ), we have

Then, is continuous and differentiable everywhere, so we may apply Rolle’s theorem on any interval. So, by Rolle’s theorem we know has at most two zeros (if there were three or more, say and , then there must be distinct numbers and with such that , but we know there is only one such that ).
Furthermore, has at least two zeros since , , and . Thus, by Bolzano’s theorem there are zeros between each of these points. We have that the number of zeros of is at most two and at least 2. Hence, the number of zeros must be exactly two

Explain why the absence of a zero does not violate Rolle’s theorem

Consider the function

Show that and , but that the derivative for all . Explain why this does not violate Rolle’s theorem.

Proof. First, we show that and by a direct computation:

Then, we compute the derivative,

To show for any we consider three cases:

• If then implies (since times a negative is positive).
• If then implies (since times a positive is then negative).
• If , then is undefined (since ).

Thus, for any

This is not a violation of Rolle’s theorem since the theorem requires that be differentiable for all on the open interval . Since is not defined at , we have is not differentiable on the whole interval. Hence, Rolle’s theorem does not apply.

Show a function satisfies the hypotheses of the mean value theorem

Let

1. Draw the graph of for .
2. Show that the hypotheses of the mean value theorem are satisfied on and find the mean values the theorem provides.

1. The sketch is as follows:

2. Since and are continuous on and , respectively and are differentiable on and , the only point at which this piecewise function might be discontinuous or non differentiable is at .

At we have

Thus, the left and right-hand limits are equal, so the limit exists and equals the function value. Therefore, is continuous at . Finally, we must check that the derivative exists at . Since the derivative of and the derivative of , both are equal to at . Hence, the derivative exists.

Now, we have met the conditions of the mean-value theorem, so we can apply the theorem to conclude,

Further, we know

So,

and

Prove a property of the integral of the product of continuous functions

Let be a continuous function on the interval and assume

for every function which is continuous on the interval . Prove that for all .

Proof. Since must hold for every function that is continuous on , it must hold for itself (since is continuous on by hypothesis). Therefore we must have,

However, for all . We know from the previous exercise (Section 3.20, #7) that a non-negative function whose integral is zero on an interval must be zero at every point at which it is continuous. By hypothesis is continuous at every point of ; hence, is also continuous at every point of (since the product of continuous functions is continuous). Therefore,

Which implies,